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M16 q 34

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M16 q 34 [#permalink] New post 26 Jul 2012, 06:02
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If a , b , c , and d are non-zero numbers such that \frac{a}{b} = \frac{c}{d} and \frac{a}{d} = \frac{b}{c} , which of the following must be true?

|a| = |c|
|b| = |d|
|a| = |d|
|b| = |a|
|b| = |c|

Because \frac{a}{b} = \frac{c}{d} , it is true that ad = bc or c = a*\frac{d}{b} . Because \frac{a}{d} = \frac{b}{c} , it is true that ac = bd or c = b*\frac{d}{a} . Thus, c = a*\frac{d}{b} = b*\frac{d}{a} . Because d is not 0, \frac{a}{b} = \frac{b}{a} or a^2 = b^2 or |a| = |b| . To see that the other choices are not necessarily true consider a = 1 , b = -1 , c = -2 , d = 2 .
The correct answer is D.

Can someone please explain me how to solve this problem efficiently? I don't understand the approach used in the solution to manipulate the variables? I did some variable manipulations but figured that if you don't do that a certain way you end up with a different answer like Mod(c) = Mod(d) which is not one of the answer choices?
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Re: M16 q 34 [#permalink] New post 26 Jul 2012, 10:00
teal wrote:
If a , b , c , and d are non-zero numbers such that \frac{a}{b} = \frac{c}{d} and \frac{a}{d} = \frac{b}{c} , which of the following must be true?

|a| = |c|
|b| = |d|
|a| = |d|
|b| = |a|
|b| = |c|

Because \frac{a}{b} = \frac{cb}{dc} , it is true that ad = bc or c = a*\frac{d}{b} . Because \frac{a}{d} = \frac{b}{c} , it is true that ac = bd or c = b*\frac{d}{a} . Thus, c = a*\frac{d}{b} = b*\frac{d}{a} . Because d is not 0, \frac{a}{b} = \frac{b}{a} or a^2 = b^2 or |a| = |b| . To see that the other choices are not necessarily true consider a = 1 , b = -1 , c = -2 , d = 2 .
The correct answer is D.

Can someone please explain me how to solve this problem efficiently? I don't understand the approach used in the solution to manipulate the variables? I did some variable manipulations but figured that if you don't do that a certain way you end up with a different answer like Mod(c) = Mod(d) which is not one of the answer choices?


The four numbers can be any non-zero real numbers. You can also use some basic algebra and manipulate the two proportions. For example:

Multiply side-by-side the two equalities. It is in fact just simple multiplication of two fractions.
You get \frac{a^2}{bd}=\frac{cb}{dc}, from which it follows that a^2=b^2 (after reduction and cross-multiplication).
Now, take the square root of both sides, and get |a|=|b|.

You can do all the manipulation above as all the numbers are non-zero. And don't forget that \sqrt{x^2}=|x|.
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Re: M16 q 34   [#permalink] 26 Jul 2012, 10:00
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